Cardano was given a word problem by another mathematician, which equated to this
equation:

x4 + 6x2
+ 36 = 60x (original equation)

Cardano was unable to solve it. He then passed it onto Ferrari who managed to solve it.
It is almost a quartic equation (ax4 + bx3 + cx2 + dx + e
= 0) except for the fact that it does not have a bx3
term in it. This makes it a depressed quartic.

Ferraris Solution to the Equation

Reference
Structure/Equation/Step

Applying
to Cardanos Equation

Description

2. x4
+ px2 + qx + r = 0

x4
+ 6x2  60x + 36 = 0

Taking 60x from both sides

2. (x2
+ p)2 = px2
 qx  r + p2

(x2
+ 6)2 = 6x2 + 60x
 36 + 36

Making a perfect square

3. (x2
+ p + y)2 = (2y + p)x2 - qx + (y2 +
12y)

(x2
+ 6 + y)2 = (2y + 6)x2
+ 60x + (y2 + 12y)

Introducing y into the perfect square. The
right-hand side has been made a quadratic equation in x, in the form y = ax2 + bx + c.

4. (-q)2
 4(p + 2y)(p2  r
+ 2py + y2) = 0

3600  4(6 + 2y)(12y + y2)
= 0

Making a perfect square with y. This is done by making a
discriminant zero*.

= 8y3
 120y2 288y  3600 = 0

Multiplying out the brackets

ax3 + bx2+ cx + d = 0

= -y3
 15y2- 36y + 450 = 0

Dividing equation by 8 to simplify. It is now a cubic
equation*. This is known as the resolvent cubic of the original quartic.

5. (x2 + 6 + y)2 = (2y
+ 6)x2 + 60x + (y2
+ 12y)

(x2 + 6 +
4)2» (2(4) + 6)x2 + 60x + (42 + 12(4))

Solved the cubic equation and found that y» 4. Now substituting into the equation in step 2.

(x2
+ 10)2» 14x2 + 60x + 64

Simplifying

(x2 + 10)2» (14x + 8)2

Making a perfect square with the right-hand side

6.

x2 + 10 »±(14x + 8)

Taking the square root of both sides

ax2 + bx + c
= 0

x2± 14x
+10 ± 8 = 0

In quadratic form (at least enough to substitute into the
quadratic formula)